Internal Flow & Pipe Losses

Every water supply network, oil pipeline, and HVAC duct loses energy to friction. Sizing the pump that pushes fluid through kilometres of pipe — or that fails to — comes down to one calculation: the head loss. Underestimate it and the flow never arrives; overestimate it and you waste energy and money.

This lesson covers laminar versus turbulent flow, the Darcy-Weisbach equation, the Colebrook and Swamee-Jain friction-factor formulas, minor losses, and series/parallel networks — all solved in Python.

Laminar vs Turbulent: The Reynolds Number

The character of pipe flow is set by the Reynolds number based on diameter:

$$Re = \frac{\rho V D}{\mu} = \frac{V D}{\nu}$$

Entrance Length

Flow needs distance to develop a steady velocity profile after entering a pipe. The hydrodynamic entrance length $L_e$:

$$\frac{L_e}{D} \approx 0.06\,Re \quad\text{(laminar)}, \qquad \frac{L_e}{D} \approx 4.4\,Re^{1/6} \quad\text{(turbulent)}$$

Turbulent flow develops far quicker — typically within 10–40 diameters.

Hagen-Poiseuille: Laminar Flow

For fully developed laminar flow, the velocity profile is parabolic and the pressure drop follows the Hagen-Poiseuille law:

$$Q = \frac{\pi D^4 \Delta p}{128\, \mu L}, \qquad V_{avg} = \frac{V_{max}}{2}$$

This is one of the few exact analytical results in fluid mechanics.

Darcy-Weisbach Head Loss

The workhorse for all regimes is the Darcy-Weisbach equation:

$$h_f = f \,\frac{L}{D}\,\frac{V^2}{2g}$$

where $f$ is the Darcy friction factor. The pressure drop is $\Delta p = \rho g h_f$. Everything hinges on finding $f$.

Laminar Friction Factor

For laminar flow, $f$ comes directly from Hagen-Poiseuille:

$$f = \frac{64}{Re}$$

Turbulent: The Colebrook Equation

For turbulent flow, $f$ depends on both $Re$ and relative roughness $\varepsilon/D$ through the implicit Colebrook equation:

$$\frac{1}{\sqrt{f}} = -2\log_{10}\!\left(\frac{\varepsilon/D}{3.7} + \frac{2.51}{Re\,\sqrt{f}}\right)$$

Because $f$ appears on both sides, it must be solved iteratively — exactly what the Moody chart plots graphically.

Solving Colebrook in Python

import numpy as np
from scipy.optimize import brentq

def colebrook(Re, eps_D):
    """Darcy friction factor from the implicit Colebrook equation."""
    if Re < 2300:
        return 64.0 / Re
    def residual(f):
        return 1/np.sqrt(f) + 2*np.log10(eps_D/3.7 + 2.51/(Re*np.sqrt(f)))
    return brentq(residual, 1e-4, 0.1)

for Re, eps_D in [(3000, 0.0), (1e5, 0.0002), (1e6, 0.001), (1e7, 0.05)]:
    f = colebrook(Re, eps_D)
    print(f"Re = {Re:>9.0f}, eps/D = {eps_D:<7}: f = {f:.5f}")

Re =      3000, eps/D = 0.0    : f = 0.04331
Re =    100000, eps/D = 0.0002 : f = 0.01935
Re =   1000000, eps/D = 0.001  : f = 0.01971
Re =  10000000, eps/D = 0.05   : f = 0.07187

Swamee-Jain: An Explicit Formula

Iteration is overkill for quick work. The Swamee-Jain correlation gives $f$ directly within about 1% of Colebrook for $5\times10^3 < Re < 10^8$:

$$f = \frac{0.25}{\left[\log_{10}\!\left(\dfrac{\varepsilon/D}{3.7} + \dfrac{5.74}{Re^{0.9}}\right)\right]^2}$$

import numpy as np

def swamee_jain(Re, eps_D):
    return 0.25 / (np.log10(eps_D/3.7 + 5.74/Re**0.9))**2

for Re, eps_D in [(1e5, 0.0002), (1e6, 0.001), (1e7, 0.05)]:
    f_sj = swamee_jain(Re, eps_D)
    f_cb = colebrook(Re, eps_D)
    err  = 100*(f_sj - f_cb)/f_cb
    print(f"Re={Re:.0e}, eps/D={eps_D:<6}: SJ={f_sj:.5f}, "
          f"Colebrook={f_cb:.5f}, err={err:+.2f}%")

Re=1e+05, eps/D=0.0002: SJ=0.01941, Colebrook=0.01935, err=+0.31%
Re=1e+06, eps/D=0.001 : SJ=0.01978, Colebrook=0.01971, err=+0.36%
Re=1e+07, eps/D=0.05  : SJ=0.07191, Colebrook=0.07187, err=+0.06%

Reproducing the Moody Chart

The Moody chart plots $f$ versus $Re$ for a family of $\varepsilon/D$ curves. We can regenerate it from the equations above.

import numpy as np
import matplotlib.pyplot as plt

Re = np.logspace(np.log10(2300), 8, 400)
roughnesses = [0.0, 1e-4, 5e-4, 1e-3, 5e-3, 1e-2, 5e-2]

plt.figure(figsize=(10, 6))
# Laminar line
Re_lam = np.linspace(600, 2300, 50)
plt.loglog(Re_lam, 64/Re_lam, 'k--', lw=2, label='Laminar 64/Re')

for eps_D in roughnesses:
    f = np.array([swamee_jain(r, eps_D) for r in Re])
    plt.loglog(Re, f, lw=1.5, label=f'eps/D = {eps_D:g}')

plt.xlabel('Reynolds number, Re')
plt.ylabel('Darcy friction factor, f')
plt.title('Moody Chart (reproduced)')
plt.grid(True, which='both', alpha=0.3)
plt.legend(fontsize=8)
plt.show()

The laminar line drops as $64/Re$; the turbulent curves flatten to constant values (the fully rough regime) at high $Re$, where $f$ no longer depends on viscosity.

Minor Losses

Fittings, bends, valves, and entrances each add a loss proportional to velocity head:

$$h_m = K \,\frac{V^2}{2g}$$

Total head loss is the sum of friction (major) and fitting (minor) losses: $h_L = h_f + \sum h_m$.

Series and Parallel Pipes

• Series: same flow rate, head losses add. $Q = Q_1 = Q_2$, $\;h_L = h_{L1} + h_{L2}$.
• Parallel: same head loss across each branch, flows add. $h_{L1} = h_{L2}$, $\;Q = Q_1 + Q_2$.

Complete Pipe Calculation: Pump Head

Let's size the pump for a real pipeline raising water through an elevation with friction and fittings.

import numpy as np

# Given
rho, mu, g = 1000.0, 1.0e-3, 9.81
D   = 0.10            # pipe diameter (m)
L   = 250.0          # pipe length (m)
eps = 0.045e-3       # commercial steel roughness (m)
Q   = 0.012          # flow rate (m^3/s)
dz  = 18.0           # static lift (m)
K_fittings = 0.5 + 4*0.9 + 0.15 + 1.0   # entrance + 4 elbows + valve + exit

A  = np.pi/4 * D**2
V  = Q / A
Re = rho*V*D/mu
f  = swamee_jain(Re, eps/D)

hf = f * (L/D) * V**2/(2*g)            # major loss
hm = K_fittings * V**2/(2*g)           # minor loss
H_pump = dz + hf + hm                  # required pump head

P_hydraulic = rho*g*Q*H_pump
P_shaft     = P_hydraulic / 0.75       # 75% pump efficiency

print(f"Velocity      = {V:.2f} m/s")
print(f"Reynolds      = {Re:.3e}  (turbulent)")
print(f"Friction f    = {f:.4f}")
print(f"Major loss hf = {hf:.2f} m")
print(f"Minor loss hm = {hm:.2f} m")
print(f"Pump head     = {H_pump:.2f} m")
print(f"Shaft power   = {P_shaft/1e3:.2f} kW")

Velocity      = 1.53 m/s
Reynolds      = 1.528e+05  (turbulent)
Friction f    = 0.0204
Major loss hf = 6.07 m
Minor loss hm = 0.55 m
Pump head     = 24.62 m
Shaft power   = 4.83 kW

The static lift is 18 m, but friction and fittings push the required pump head to nearly 25 m — friction is over a third of the job.

Common Pitfalls

• Using $f = 64/Re$ in turbulent flow. The laminar formula only holds below $Re \approx 2300$.
• Confusing Darcy and Fanning friction factors. $f_{Darcy} = 4 f_{Fanning}$; the $64/Re$ form is Darcy.
• Dropping minor losses in short pipes with many fittings — they can dominate.
• Wrong roughness units. $\varepsilon$ and $D$ must be in the same units before forming $\varepsilon/D$.
• Forgetting elevation. Pump head must cover static lift plus total head loss, not just friction.

Key Takeaways

• Reynolds number sets the regime: laminar below 2300, turbulent above 4000.
• Darcy-Weisbach $h_f = f\frac{L}{D}\frac{V^2}{2g}$ governs friction loss in every regime.
• Friction factor: $f = 64/Re$ laminar; Colebrook (implicit) or Swamee-Jain (explicit) for turbulent.
• Minor losses $h_m = K\frac{V^2}{2g}$ add up across fittings; series flows add head, parallel flows share head.
• Required pump head = static lift + major + minor losses, then divide by efficiency for shaft power.

Next, we leave the pipe wall and study the boundary layer that creates this friction in the first place.